c++ - 为什么在使用大括号初始值设定项列表时首选 std::initializer_list 构造函数？

Question

考虑代码

#include <iostream>

class Foo
{
    int val_;
public:
    Foo(std::initializer_list<Foo> il)
    {
        std::cout << "initializer_list ctor" << std::endl;
    }
    /* explicit */ Foo(int val): val_(val)
    {
        std::cout << "ctor" << std::endl;
    };
};

int main(int argc, char const *argv[])
{
    // why is the initializer_list ctor invoked?
    Foo foo {10}; 
}

输出是

ctor
initializer_list ctor

据我了解，值 10 被隐式转换为 Foo (第一个 ctor 输出)，然后初始化构造函数启动在(第二个 initializer_list ctor 输出)。我的问题是为什么会这样？标准构造函数 Foo(int) 不是更好的匹配吗？即，我希望这个片段的输出只是 ctor。

PS:如果我将构造函数 Foo(int) 标记为 explicit，那么 Foo(int) 是唯一调用的构造函数，如整数 10 现在不能隐式转换为 Foo。

Answer 1

§13.3.1.7 [over.match.list]/p1:

When objects of non-aggregate class type T are list-initialized (8.5.4), overload resolution selects the constructor in two phases:

• Initially, the candidate functions are the initializer-list constructors (8.5.4) of the class T and the argument list consists of the initializer list as a single argument.
• If no viable initializer-list constructor is found, overload resolution is performed again, where the candidate functions are all the constructors of the class T and the argument list consists of the elements of the initializer list.

If the initializer list has no elements and T has a default constructor, the first phase is omitted. In copy-list-initialization, if an explicit constructor is chosen, the initialization is ill-formed.

只要有一个可行的初始化列表构造函数，当使用列表初始化并且初始化列表至少有一个元素时，它将胜过所有非初始化列表构造函数。